Stack以及ArrayDeque、LinkedList的区别练习

Stack类

可以看你到stack的源码很少，它是基于Vector集合来实现的，而Vector集合本来就是线程安全的，使用Synchronize的方式实现的，效率比较低，所以不建议使用。

public
class Stack<E> extends Vector<E> {
    /**
     * Creates an empty Stack.
     */
    public Stack() {
    }

    /**
     * Pushes an item onto the top of this stack. This has exactly
     * the same effect as:
     * <blockquote><pre>
     * addElement(item)</pre></blockquote>
     *
     * @param   item   the item to be pushed onto this stack.
     * @return  the <code>item</code> argument.
     * @see     java.util.Vector#addElement
     */
    public E push(E item) {
        addElement(item);

        return item;
    }

    /**
     * Removes the object at the top of this stack and returns that
     * object as the value of this function.
     *
     * @return  The object at the top of this stack (the last item
     *          of the <tt>Vector</tt> object).
     * @throws  EmptyStackException  if this stack is empty.
     */
    public synchronized E pop() {
        E       obj;
        int     len = size();

        obj = peek();
        removeElementAt(len - 1);

        return obj;
    }

    /**
     * Looks at the object at the top of this stack without removing it
     * from the stack.
     *
     * @return  the object at the top of this stack (the last item
     *          of the <tt>Vector</tt> object).
     * @throws  EmptyStackException  if this stack is empty.
     */
    public synchronized E peek() {
        int     len = size();

        if (len == 0)
            throw new EmptyStackException();
        return elementAt(len - 1);
    }

    /**
     * Tests if this stack is empty.
     *
     * @return  <code>true</code> if and only if this stack contains
     *          no items; <code>false</code> otherwise.
     */
    public boolean empty() {
        return size() == 0;
    }

    /**
     * Returns the 1-based position where an object is on this stack.
     * If the object <tt>o</tt> occurs as an item in this stack, this
     * method returns the distance from the top of the stack of the
     * occurrence nearest the top of the stack; the topmost item on the
     * stack is considered to be at distance <tt>1</tt>. The <tt>equals</tt>
     * method is used to compare <tt>o</tt> to the
     * items in this stack.
     *
     * @param   o   the desired object.
     * @return  the 1-based position from the top of the stack where
     *          the object is located; the return value <code>-1</code>
     *          indicates that the object is not on the stack.
     */
    public synchronized int search(Object o) {
        int i = lastIndexOf(o);

        if (i >= 0) {
            return size() - i;
        }
        return -1;
    }

    /** use serialVersionUID from JDK 1.0.2 for interoperability */
    private static final long serialVersionUID = 1224463164541339165L;

可以使用双端队列代替Stack，效果更好（参考）

使用方式

public static void main(String[] args) throws IOException {
        Stack<Integer> stack = new Stack<>();
        stack.push(1);
        ArrayDeque<Integer> deque = new ArrayDeque<>();
        deque.push(1);
        deque.push(2);
        deque.push(3);
        deque.push(4);
        deque.offerFirst(5);
        deque.offerFirst(6);
        deque.push(7);
        deque.push(8);
        while (!deque.isEmpty()){
            System.out.println(deque.pop());
        }
        //8 7 6 5 4 3 2 1
    }
public static void main(String[] args) throws IOException {
        Stack<Integer> stack = new Stack<>();
        stack.push(1);
        ArrayDeque<Integer> deque = new ArrayDeque<>();
        deque.push(1);
        deque.push(2);
        deque.push(3);
        deque.push(4);
        deque.offerLast(5);
        deque.offerLast(6);
        deque.push(7);
        deque.push(8);
        while (!deque.isEmpty()){
            System.out.println(deque.pop());
        }
        //8 7 4 3 2 1 5 6
    }

ArrayDeque底层通过数组实现，为了满足可以同时在数组两端插入或删除元素的需求，该数组还必须是循环的，即循环数组（circular array），也就是说数组的任何一点都可能被看作起点或者终点。ArrayDeque是非线程安全的（not thread-safe），当多个线程同时使用的时候，需要程序员手动同步；另外，该容器不允许放入null元素。

双向链表实现双端队列

LinkedList<Integer> list = new LinkedList<>();
        list.push(1);
        list.push(2);
        list.push(3);
        list.push(4);
        list.offerLast(5);
        list.offerLast(6);
        list.push(7);
        list.push(8);
        while (!list.isEmpty()){
            System.out.println(list.pop());
        }

由于 LinkedList 基于链表实现，存储元素过程中，无需像 ArrayList 那样进行扩容。但有得必有失，LinkedList 存储元素的节点需要额外的空间存储前驱和后继的引用。另一方面，LinkedList 在链表头部和尾部插入效率比较高，但在指定位置进行插入时，效率一般。原因是，在指定位置插入需要定位到该位置处的节点，此操作的时间复杂度为 O(N)。
同时，LinkedList也是非线程安全的，并发环境下，多个线程同时操作 LinkedList，会引发不可预知的错误。